Electronic – Is it right to apply superposition theorem to the RC circuit with non-zero initial capacitor voltage

capacitorcircuit analysis

I'm having confusion with applying superposition to simple RC circuit with non-zero initial capacitor voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

Consider a simple RC circuit series connected with voltage source of step voltage \$V_s=8u(t)\$, and let's say initial voltage of the capacitor is \$V_c = 2V\$.

It's obvious that the value of \$V_c\$ over time is

\$
V_c = 2+(8-2)(1-e^{t/RC})
\$.

But if I split the voltage source \$V_s=8u(t)\$ with two voltage sources \$V_{s1}=4u(t), V_{s2}=4u(t)\$, then the corresponding \$V_{c1}\$ and \$V_{c2}\$ will be

\$
V_{c1} = 2+(4-2)(1-e^{t/RC})
\$.

\$
V_{c2} = 2+(4-2)(1-e^{t/RC})
\$, respectively.

Then, applying superposition yields \$V_{c1}+V_{c2}=4+4(1-e^{t/RC})\$, which is different from

\$
V_c = 2+(8-2)(1-e^{t/RC})
\$.

Where am I doing wrong?

Is it wrong to apply superposition theorem to non-zero state of capacitor?

Best Answer

Yes, you can use superposition -- once for the V source, and once for the cap decay.

The cap decay is (set the other V to 0) Vc=2+(0−2)(1−e(-t/RC)), or 2.e(-t/RC) [you missed a '-' in the exponential]

The 8 V source adds Vc += 8(1-e(-t/RC))

Two 4 V sources would each add Vc += 4(1-(e-t/RC)).

They all add together as expected.